Matlab Error after executing the code

Question

venkatesu u 2021-11-4

0
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此问题的直接链接

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回答： VBBV 2021-11-30

Dear sir/Madam

I have the following Matlab code.

clear all

clc

kc=[1 2 3 4 5 6 7 8 9 0];

k12=200;

k34=300;

new1=1:1:10;

for h1=1:length(new1);

for l1=1:length(new1);

K(h1,l1)=[k12 -k12 0 0;-k12 kc(h1,l1)+k12 -kc(h1,l1) 0;0 -kc(h1,l1) k34+kc(h1,l1) -k34;0 0 -k34 k34];

end

After executing it, an error is displayed as "Unable to perform assignment because the left and right sides have a different number of elements." Kindly provide any suggestions to avoid this error.

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Sreedevi K 2021-11-4

You are trying to save a 4*4 matrix (RHS) into a single element (LHS) which is not allowed.

venkatesu u 2021-11-4

Dear Madam I have tried your code in MATLAB but it shows following error "Index in position 1 exceeds array bounds. Index must not exceed 1." Kindly provide suggestions to avoid error

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Answer 1

Sreedevi K 2021-11-4

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You have to save into a cell array or a 3D matrix. As shown below:

kc=[1 2 3 4 5 6 7 8 9 0];
k12=200;
k34=300;
new1=1:1:10;
count = 0 ; 
for h1=1:length(new1)
    for l1=1:length(new1)
        count = count+1 ;
        K{count}=[k12 -k12 0 0;-k12 kc(h1,l1)+k12 -kc(h1,l1) 0;0 -kc(h1,l1) k34+kc(h1,l1) -k34;0 0 -k34 k34];
    end
end

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Answer 2

VBBV 2021-11-30

0
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https://ww2.mathworks.cn/matlabcentral/answers/1578700-matlab-error-after-executing-the-code#answer_843434

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clear all
clc
kc=[1 2 3 4 5 6 7 8 9 0];
k12=200;
k34=300;
new1=1:1:10;
for l1=1:length(new1)
     K(:,:,:,:,l1)=[k12 -k12 0 0;-k12 kc(l1)+k12 -kc(l1) 0;0 -kc(l1) k34+kc(l1) -k34;0 0 -k34 k34];
end
K
K = 
K(:,:,1,1,1) =

   200  -200     0     0
  -200   201    -1     0
     0    -1   301  -300
     0     0  -300   300


K(:,:,1,1,2) =

   200  -200     0     0
  -200   202    -2     0
     0    -2   302  -300
     0     0  -300   300


K(:,:,1,1,3) =

   200  -200     0     0
  -200   203    -3     0
     0    -3   303  -300
     0     0  -300   300


K(:,:,1,1,4) =

   200  -200     0     0
  -200   204    -4     0
     0    -4   304  -300
     0     0  -300   300


K(:,:,1,1,5) =

   200  -200     0     0
  -200   205    -5     0
     0    -5   305  -300
     0     0  -300   300


K(:,:,1,1,6) =

   200  -200     0     0
  -200   206    -6     0
     0    -6   306  -300
     0     0  -300   300


K(:,:,1,1,7) =

   200  -200     0     0
  -200   207    -7     0
     0    -7   307  -300
     0     0  -300   300


K(:,:,1,1,8) =

   200  -200     0     0
  -200   208    -8     0
     0    -8   308  -300
     0     0  -300   300


K(:,:,1,1,9) =

   200  -200     0     0
  -200   209    -9     0
     0    -9   309  -300
     0     0  -300   300


K(:,:,1,1,10) =

   200  -200     0     0
  -200   200     0     0
     0     0   300  -300
     0     0  -300   300
clear K 
TT = 1
TT = 1
for l1=1:length(new1)
     K{TT}=[k12 -k12 0 0;-k12 kc(l1)+k12 -kc(l1) 0;0 -kc(l1) k34+kc(l1) -k34;0 0 -k34 k34];     
     TT =  TT+1;     
end
K
K = 1×10 cell array
    {4×4 double}    {4×4 double}    {4×4 double}    {4×4 double}    {4×4 double}    {4×4 double}    {4×4 double}    {4×4 double}    {4×4 double}    {4×4 double}

You need one for loop only to compute the matrix, Also, you were getting error because, when you access more elements from matrix than its actual dimension. You can do it using cell array as below